BGC, EHF: the greater BAC is equal to the greater EDF, and the less BGC to the less EHF. Take (1. 3.) K, L, the centres of the circles, and join BK, KC, EL, LF; and because the circles are equal, the straight lines from their centres are equal; therefore BK, KC are equal to EL, LF; but the base BC is also equal to the base EF; therefore the angle BKC is equal (8. 1.) to the angle ELF: and equal angles stand upon equal (26.3.) arcs, when they are at the centres; therefore the arc BGC is equal to the arc EHF. But the whole circle ABC is equal to the whole EDF; the remaining part, therefore, of the circumference viz. BAC, is equal to the remaining part EDF. PROP. XXIX. THEOR. In equal circles equal arcs are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the arcs BGC, EHF also be equal; and join BC, EF: the straight line BC is equal to the straight line EF. Take (1.3.) K, L the centres of the circles, and join BK, KC, EL, LF : and because the arc BGC is equal to the arc EHF, the angle BKC is equal (27. 3.) to the angle ELF: also because the circles ABC, DEF are equal, their radii are equal: therefore BK, KC are equal to EL, LF: and they contain equal angles; therefore the base BC is equal (4. 1.) to the base EF. PROP. XXX. THEOR. To bisect a given arc, that is, to divide it into two equal parts. Let ADB be the given arc; it is required to bisect it. Join AB, and bisect (10. 1.) it in C; from the point C draw CD at right angles to AB, and join AD, DB: the arc ADB is bisected in the point D. Because AC is equal to CB, and CD common to the triangle ACD, BCD, the two sides AC, CD are equal to the D two BC, CD; and the angle ACD is equal to the angle BCD, because each of them is a right angle: therefore the base AD is equal (4. 1.) to the base BD. But equal straight lines cut off equal arcs, (28. 3.) the greater equal to the greater, and the less to the less; and AD, DB are each of them less than a semicircle, because DC passes through the centre (Cor. 1.3.); wherefore the arc AD is equal to the arc DB: and therefore the given arc ADB is bisected in D. SCHOLIUM. By the same construction, each of the halves AD, DB may be divided into two equal parts; and thus, by successive subdivisions, a given arc may be divided into four, eight, sixteen, &c. equal parts. PROP. XXXI. THEOR. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and centre E; draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle. F A D Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal (5. 1.) to EBA: also because AE is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB. But FAC, the exterior angle of the triangle ABC, is also equal (32. 1.) to the two angles ABC, ACB; therefore the an gle BAC is equal to the angle FAC, and B each of them is therefore a right angle (7. def. 1.); wherefore the angle BAC in a semicircle is a right angle. C E And because the two angles ABC, BAC of the triangle ABC are together less (17.1.) than two right angles, and BAC is a right angle, ABC must be less than a right angle; and therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle. Also because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (22. 3.) to two right angles; therefore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other ADC is greater than a right angle. Cor. From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are equal, they are right angles. PROP. XXXII. THEOR. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line which touches the circle, shall be equal to the angles in the alternate segments of the circle. Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn cutting the circle: the angles which BD makes with the touching line EF shall be equal to the angles in the alternate segments of the circle: that is, the angle FBD is equal to 1 the angle which is in the segment DAB, and the angle DBE to the angle in the segment BCD. From the point B draw (11. 1.) BA at right angles to EF, and take any point C in the arc BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line, from the point of contact B, the centre of the circle is (19.3.) in BA; therefore the an- A D C E B F 1 11 PROP. XXXIII. PROB. Upon a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle, Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C. First, let the angle at C be a right angle; bisect (10.1.) AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; the an AB, make (23.1.) the angle BAD equal to the angle C, and from the point A draw (11. 1.) AE at right angles to AD; bisect (10.1.) AB in F, and H from F draw (11.1.) FG at right angles to AB, and join GB: then the circle described from the centre G, at the distance GA, shall pass D through the point B; let this be the circle AHB: and because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD (Cor. 1. 16. 3.) touches H the circle; and because AB, drawn from C the point of contact A, cuts the circle, F the angle DAB is equal to the angle in A B the alternate segment AHB (32. 3.); but the angle DAB is equal to the angle G C, therefore also the angle C is equal to the angle in the segment AHB: Where E fore, upon the given straight line AB the segment AHB of a circle is describ D ed which contains an angle equal to the given angle at C. To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the angle D. Draw (17.3.) the straight line EF touching the circle ABC in the point B, and at the point B, in the straight line BF make (23.1.) the angle FBC A equal to the angle D; therefore, be cause the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal (32. 3.) to the an C gle in the alternate segment BAC; segment BAC is equal to the angle D: wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E; the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED. If AC, BD pass each of them through the centre, so that E is the centre, it is evident that AE, EC, BE, ED, being all equal, the rectangle AE. A EC is likewise equal to the rectangle BE.ED. But let one of them BD pass through the cen- B tre, and cut the other AC, which does not pass through the centre, at right angles in the point E; then, if BD be bisected in F, F is the centre of C the circle ABCD; join AF: and because BD, which passes through the centre, cuts the straight line AC, which does not D pass through the centre at right angles, in E, AE, EC are equal (3.3.) to one another; and because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, BE.ED (5. 2.) + EF2 = FB2 = AF2. But AF2 = AE2 + (47.1.) EF2, therefore BE.ED + F EF2, = AE2 + EF2, and taking EF2 from each, A BE.ED=AE2-AE.EC. Next, let BD, which passes through the centre, cut the other AC, which does not pass through E C B |